All categories

3 years ago

Need QUICK answer!

Physics

2 answers:

Tresset [83]3 years ago

7 0

Most likely the answer is B.

Oduvanchick [21]3 years ago

6 0

I am pretty sure it is D i would check but

You might be interested in

Often called velocity this is the velocity of an object at a particular moment in time

Yuliya22 [10]

Instantaneous speed?

8 0

2 years ago

A street light is on top of a 8 foot pole. Joe,

zubka84 [21]

8/4 = y/y-x

8y - 8x = 4y

y = 2x

y = 2 x 4

y = 8

Hope this helps

5 0

3 years ago

A 2.74 g coin, which has zero potential energy at the surface, is dropped into a 12.2 m well. After the coin comes to a stop in

VikaD [51]

Answer:

B. - 0.328

Explanation

Potential Energy: This is the energy of a body due to position.

The S.I unit of potential energy is Joules (J).

It can be expressed mathematically as

Ep = mgh........................... Equation 1

Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,

Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²

Substituting these values into equation 1

Ep = 0.00274×12.2×9.8

Ep = 0.328 J.

Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J

The right option is B. - 0.328

7 0

2 years ago

Calculate the approximate volume of a uranium nucleus, 23592u. (you can ignore the mass defect in this calculation and simply ta

tester [92]

Radius of nuclei is given by formula

$R = R_oA^{1/3}$

now we can say volume of the nuclei is given as

$V = \frac{4}{3}\pi R_o^3* A$

now the density is given as

density = mass / volume

mass of nuclei = mass of neutron + mass of protons

$m = z*m_p + (A- z)*m_n$

$m_p = m_n = 1.008u$

$m = A*1.008u$

Now density is given as

$\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}$

here we know that

$R_0$ = 1.2 fm

$\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}$

$\rho = 2.31 * 10^{17} kg/m^3$

So from above we can say that density of all nuclei is almost same.

5 0

3 years ago

Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0

DENIUS [597]

Answer:

Tension in the string is equal to 58.33 N ( this will be the strength of the string )

Explanation:

We have given mass m = 1.7 kg

radius of the circle r = 0.48 m $F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N$

Kinetic energy is given 14 J

Kinetic energy is equal to $KE=\frac{1}{2}mv^2$

So $\frac{1}{2}\times 1.7\times v^2=14$

$v^2=16.47$

v = 4.05 m/sec

Centripetal force is equal to $F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N$

So tension in the string will be equal to 58.33 N ( this will be the strength of the string )

3 0

3 years ago